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CS300
1)
ω |
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| R | 2 | |
= | - ( | ) | ||||
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| L * C | 2 * L | ||||
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Having as a unit of measure the “L” inductance in Henry, the “C” capacitor in Farad and the “R” resistance in Ohm, according to the above mentioned data:
| ω | = 1331.21 rad/S | ||||
2) |
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| α | = |
| R | ||
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| 2 * L | |||||
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from which: | α | = 357.14 | ||||
3) |
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| t M = | π | ||||
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| 2 * ω | |||||
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from which: | tM = 0.00117 s |
( tM states the time needed by the current to reach its maximum value )
4)the peak current can be calculated with the following formula:
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| V | α * t M |
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I P = ( | ω | * L | )* e |
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from which : | IP = 572.3A |
It is obvious that considering a 70V discharge of the
IP= 286.1 A
Such value meets the needing of both the converter (which for short periods is able to bear a current value two times the rated one) and the inductance, whose saturation current is higher than 300A.
Table of
17 |